\(\int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx\) [121]

Optimal result

Integrand size = 21, antiderivative size = 54 \[ \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \sqrt {b \sqrt {x}+a x}}{3 b x}+\frac {8 a \sqrt {b \sqrt {x}+a x}}{3 b^2 \sqrt {x}} \]

[Out]

-4/3*(b*x^(1/2)+a*x)^(1/2)/b/x+8/3*a*(b*x^(1/2)+a*x)^(1/2)/b^2/x^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2041, 2039} \[ \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx=\frac {8 a \sqrt {a x+b \sqrt {x}}}{3 b^2 \sqrt {x}}-\frac {4 \sqrt {a x+b \sqrt {x}}}{3 b x} \]

[In]

Int[1/(x^(3/2)*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x])/(3*b*x) + (8*a*Sqrt[b*Sqrt[x] + a*x])/(3*b^2*Sqrt[x])

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = -\frac {4 \sqrt {b \sqrt {x}+a x}}{3 b x}-\frac {(2 a) \int \frac {1}{x \sqrt {b \sqrt {x}+a x}} \, dx}{3 b} \\ & = -\frac {4 \sqrt {b \sqrt {x}+a x}}{3 b x}+\frac {8 a \sqrt {b \sqrt {x}+a x}}{3 b^2 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \left (b-2 a \sqrt {x}\right ) \sqrt {b \sqrt {x}+a x}}{3 b^2 x} \]

[In]

Integrate[1/(x^(3/2)*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

(-4*(b - 2*a*Sqrt[x])*Sqrt[b*Sqrt[x] + a*x])/(3*b^2*x)

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(-\frac {4 \sqrt {b \sqrt {x}+a x}}{3 b x}+\frac {8 a \sqrt {b \sqrt {x}+a x}}{3 b^{2} \sqrt {x}}\)	\(41\)
default	\(-\frac {\sqrt {b \sqrt {x}+a x}\, \left (6 x^{\frac {5}{2}} \sqrt {b \sqrt {x}+a x}\, a^{\frac {5}{2}}+6 x^{\frac {5}{2}} a^{\frac {5}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}+3 x^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+2 a \sqrt {x}+b}{2 \sqrt {a}}\right ) a^{2} b -3 x^{\frac {5}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{2} b -12 x^{\frac {3}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {3}{2}}+4 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} b \sqrt {a}\, x \right )}{3 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b^{3} x^{\frac {5}{2}} \sqrt {a}}\)	\(194\)

[In]

int(1/x^(3/2)/(b*x^(1/2)+a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-4/3*(b*x^(1/2)+a*x)^(1/2)/b/x+8/3*a*(b*x^(1/2)+a*x)^(1/2)/b^2/x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx=\frac {4 \, \sqrt {a x + b \sqrt {x}} {\left (2 \, a \sqrt {x} - b\right )}}{3 \, b^{2} x} \]

[In]

integrate(1/x^(3/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="fricas")

[Out]

4/3*sqrt(a*x + b*sqrt(x))*(2*a*sqrt(x) - b)/(b^2*x)

Sympy [F]

\[ \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx=\int \frac {1}{x^{\frac {3}{2}} \sqrt {a x + b \sqrt {x}}}\, dx \]

[In]

integrate(1/x**(3/2)/(b*x**(1/2)+a*x)**(1/2),x)

[Out]

Integral(1/(x**(3/2)*sqrt(a*x + b*sqrt(x))), x)

Maxima [F]

\[ \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b \sqrt {x}} x^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/x^(3/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(a*x + b*sqrt(x))*x^(3/2)), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx=\frac {4 \, {\left (3 \, \sqrt {a} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + b\right )}}{3 \, {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{3}} \]

[In]

integrate(1/x^(3/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="giac")

[Out]

4/3*(3*sqrt(a)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) + b)/(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^3

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx=\int \frac {1}{x^{3/2}\,\sqrt {a\,x+b\,\sqrt {x}}} \,d x \]

[In]

int(1/(x^(3/2)*(a*x + b*x^(1/2))^(1/2)),x)

[Out]

int(1/(x^(3/2)*(a*x + b*x^(1/2))^(1/2)), x)